Register to reply 
Computing Matrix, finding kernel and image 
Share this thread: 
#1
Nov210, 11:22 AM

P: 19

Let T: R[x]_{2}[tex]\rightarrow[/tex] R[x]_{3} be defined by T(P(x))=xP(x). Compute the matrix of x with respect to bases {1,x,x^{2}} and {1,x,x^{2},x^{3}}. Find the kernel and image of T.
I know how to do this when given bases without exponents, however I do not know exactly what this is saying and therefore am having a hard time starting it. 


#2
Nov210, 11:59 AM

Mentor
P: 18,213

Let's compute the first column.
You take the first basis element, that is 1. Now you'll have to express T(1)=x in terms of the basis {1,x,x²,x³}. That would be (0,1,0,0). So the first column would consist out of [tex] \left(\begin{array}{ccc} 0 & ? & ? \\ 1 & ? & ? \\ 0 & ? & ? \\ 0 & ? & ? \end{array} \right) [/tex] Now for the second and third column, you'll have to express T(x) and T(x²) in terms of the basis {1,x,x²,x³}. 


#3
Nov210, 03:15 PM

P: 19

Okay so matrix T(x)=x^{2} and with respect to the basis {1,x,x²,x³} the second column would be (0,0,1,0), T(x^{2})=x^{3} and with respect to the basis {1,x,x²,x³} would be (0,0,0,1).
I would then say that T=(0 0 0) (1 0 0) (0 1 0) (0 0 1) with respect to {1,x,x²,x³} right? Not with respect to {1,x,x²} and {1,x,x²,x³} so then to find the im(T) I just use (0 0 0) (1 0 0) (0 1 0) (0 0 1) and solve it with respect to the basis {1,x,x²,x³} right? Then how would I go about solving for the ker(T)? 


#4
Nov210, 03:24 PM

Mentor
P: 18,213

Computing Matrix, finding kernel and image
For the Ker(T). Take your matrix A. Then wonder when Ax=0. This you should be able to solve easily (it's a system with 3 unknowns and 4 equations)



#5
Nov210, 03:34 PM

Mentor
P: 18,213

So, if (x,y,z) is in the kernel, then you must have
[tex] \left( \begin{array}{ccc} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} x\\ y\\ z \end{array} \right) = 0 [/tex] It's not hard to see that this can only be the case iff x=y=z=0 


#6
Nov210, 03:45 PM

P: 19

right, that just seemed way too easy I thought I was doing it wrong.
Then the im(T) is the span of e_{2}, e_{3}, and e_{4} 


Register to reply 
Related Discussions  
Matrix Image and Kernel  Calculus & Beyond Homework  4  
Basis for Image and Kernel of matrix  Calculus & Beyond Homework  3  
Finding a Matrix whose kernel is spanned by 2 vectors  Calculus & Beyond Homework  3  
Finding kernel of matrix transformation  Calculus & Beyond Homework  2  
Kernel and image of a matrix A  Calculus & Beyond Homework  4 